Chapter 6.5, Problem 45E

To determine

To calculate: The range of values of side a of a triangle ABC such that it has two solutions, one solutions and no solution provided $\angle A=30°\text{ and }b=100$ . Sketch the triangles to verify the criteria for number of solutions provided $\angle A$ and sides a and b .

  $\begin{array}{cc}\text{Criterion}& \text{Number of solutions}\\ a\ge b& 1\\ b>a>b\sin A& 2\\ a=b\sin A& 1\\ a<b\sin A& 0\end{array}$

Answer to Problem 45E

The range of values of side a of a triangle ABC is provided below,

  $\begin{array}{cc}\text{Number of solutions}& \text{Range of }a\\ 0& a<50\\ 1& a\ge 100\\ 1& a=50\\ 2& 50<a<100\end{array}$

Explanation of Solution

Given information:

The criteria for number of solutions provided $\angle A$ and sides a and b of a triangle ABC .

  $\begin{array}{cc}\text{Criterion}& \text{Number of solutions}\\ a\ge b& 1\\ b>a>b\sin A& 2\\ a=b\sin A& 1\\ a<b\sin A& 0\end{array}$

Formula used:

The table for number of solutions,

  $\begin{array}{cc}\text{Criterion}& \text{Number of solutions}\\ a\ge b& 1\\ b>a>b\sin A& 2\\ a=b\sin A& 1\\ a<b\sin A& 0\end{array}$

Calculation:

Consider a triangle ABC with $\angle A,a\text{ and }b$ provided, so following cases arise:

Case I: When $a<b$ ,

  

The side a is very short to complete the triangle and reach the base of it. As observed no triangle exists, so there is no solution.

Case II: When $a<b$ ,

  

The side a is perpendicular to the base of the triangle. As observed it is a right angled triangle and one triangle is possible, so there is one solution.

Case III: When $a<b$ ,

  

The side a touches the base of triangle at two points. As observed there are two triangles so there are two solutions possible.

Case IV: When $a=b$ ,

  

The side a touches the base of the triangle at a single point. As observed one triangle is possible, so there is one solution.

Consider criteria for number of solutions provided $\angle A$ and sides a and b of a triangle ABC .

  $\begin{array}{cc}\text{Criterion}& \text{Number of solutions}\\ a\ge b& 1\\ b>a>b\sin A& 2\\ a=b\sin A& 1\\ a<b\sin A& 0\end{array}$

It is provided that $\angle A=30°\text{ and }b=100$ . Therefore,

  $\begin{array}{c}b\sin A=100\sin 30°\\ =100\left(\frac{1}{2}\right)\\ =50\end{array}$

The triangle ABC has no solution if $a<b\sin A$ that is,

  $\begin{array}{c}a<b\sin A\\ a<50\end{array}$

The triangle ABC has one solution if $a\ge b$ that is,

  $\begin{array}{c}a\ge b\\ a\ge 100\end{array}$

Also the triangle ABC has one solution if $a=b\sin A$ that is,

  $\begin{array}{c}a=b\sin A\\ a=50\end{array}$

The triangle ABC has two solutions if $b>a>b\sin A$ that is,

  $\begin{array}{c}b>a>b\sin A\\ 100>a>50\end{array}$

The range of values of side a of a triangle ABC is provided below,

  $\begin{array}{cc}\text{Number of solutions}& \text{Range of }a\\ 0& a<50\\ 1& a\ge 100\\ 1& a=50\\ 2& 50<a<100\end{array}$